The given sequence is $\sqrt{7}$,$\sqrt{7+ \sqrt{7}}$,$\sqrt{7+\sqrt{7+\sqrt{7}}}$,.....and so on.
the sequence is increasing so to converge must be bounded above.Now looks like they would not exceed 7. The given options are
${1+\sqrt{33}}\over{2}$
${1+\sqrt{32}}\over{2}$
${1+\sqrt{30}}\over{2}$
${1+\sqrt{29}}\over{2}$
How to proceed now. Thanks for any help.
Trick: Let $X = \sqrt{ 7 + \sqrt{ 7 + ... } } $. We have $X = \sqrt{ 7 + X } $ and so $X^2 = 7 + X $. Now you solve the quadratic equation.
Here's a quick proof that the sequence converges.
The sequence is given by the recursive formula $a_0 = 0$, and $a_{n+1} = \sqrt{7 + a_n}$.
Using the method suggested by the previous poster, let $L$ be the positive solution to the equation $x^2 = x + 7$.
We can prove that $a_n < L$ for all $L$ by induction. It is clear that $a_0 = 0 < L$. Then if $a_n < L$ we have $a_{n+1} = \sqrt{7 + a_n} < \sqrt{7 + L} = L$.
We can also prove that the sequence is increasing. Since $a_1 = \sqrt{7}$, we have $a_1 > a_0$. Now, if $a_n > a_{n-1}$, we have $a_{n+1} = \sqrt{7 + a_{n}} > \sqrt{7 + a_{n-1}} = a_{n}$.
